A) zero
B) 10%
C) 5%
D) 0.1%
Correct Answer: B
Solution :
Key Idea: The perceived frequency of source depends on the relative motion between the source and the observer. The perceived frequency of source observed by the observer is given by \[n=n\left( \frac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right)\] As sound source is stationary and observer is moving toward the source, we conclude \[n=n\left( \frac{v+{{v}_{o}}}{v} \right)\] \[(\because \,{{v}_{s}}=0)\] Given, \[{{v}_{o}}=\frac{v}{10}\] \[\therefore \] \[n=n\left( \frac{v+v/10}{v} \right)\] or \[\frac{n}{n}=\frac{11}{10}\] Hence, apparent increase in frequency is given by \[\frac{n-n}{n}=\frac{11-10}{10}=0.1=10%\]
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